How to show two identical integrals $\; 2\int_{-r}^{r}f(x)dx=r\int_{-r}^{r}\sqrt{1+(df/dx)^{2}}dx$ With
$$ \;f(x)=(r^2-x^2)^{1/2}$$
show that$$\; 2\int_{-r}^{r}f(x)dx=r\int_{-r}^{r}\sqrt{1+(df/dx)^{2}}dx$$
by using integration by parts. Then, conclude that the area of a circle of radius r is $\;\pi r^2$ if we define $\;\pi=\;$ circumference$\;\div\;$ diameter
I thought the only method I can use is via trig substitutions. But, the problem asks to solve it with integration-by-parts.
I need help to solve this problem.
Thank you.
 A: Integrate by parts as follows
\begin{align}
LHS= &\> 2\int_{-r}^{r}\sqrt{r^2-x^2} dx
=\int_{-r}^{r}\frac{\sqrt{r^2-x^2}}xd(x^2)\\
= & \>x \sqrt{r^2-x^2}\bigg|_{-r}^{r} + \int_{-r}^{r}\frac{r^2}{\sqrt{r^2-x^2}}dx\\
=&\>0+\>r\int_{-r}^{r}\sqrt{1+\left(\frac{df }{dx}\right)^{2}}dx=RHS
\end{align}
A: The equation claims the area of the centre-$O$ radius-$r$ circle is $r$ times the half/circumference due to its upper semicircle, i.e. $\pi r^2=r\cdot\tfrac12\cdot2\pi r$, which is trivial if you know the semicircle's area is $\tfrac12\pi r^2$. The challenge at hand is to prove this by IBP instead. First write the problem as$$\int_{-r}^r2\sqrt{r^2-x^2}dx=\int_{-r}^r\frac{r^2}{\sqrt{r^2-x^2}}dx.$$Now integrate$$2\sqrt{r^2-x^2}=\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}+\frac{r^2}{\sqrt{r^2-x^2}}=\left(x\sqrt{r^2-x^2}\right)^\prime+\frac{r^2}{\sqrt{r^2-x^2}}.$$
A: $$f(x)=(r^2-x^2)^{1/2}$$
$$f'(x)=-x(r^2-x^2)^{-1/2}$$
$$f'^2(x)=\frac{x^2}{r^2-x^2}$$
$$1+f'^2(x)=1+\frac{x^2}{r^2-x^2}=\frac{r^2-x^2+x^2}{r^2-x^2}=\frac{r^2}{r^2-x^2}$$
$$\sqrt{1+f'^2(x)}=r(r^2-x^2)^{-1/2}$$
so:
$$r\int_{-r}^r\sqrt{1+f'^2(x)}dx=r^2\int_{-r}^r(r^2-x^2)^{-1/2}dx$$
and:
$$2\int_{-r}^rf(x)dx=2\int_{-r}^r(r^2-x^2)^{1/2}$$
can u finish it from here?
